EV for a D'Alembert

Quote: DeMango

How about betting banker with a better than 50% win rate? Also at a casino that only charges 4% comission.

Quote: ThatDonGuy

Why would I say that? In fact, didn't I (and a few others) say that Martingale also always wins on a 50/50 game if you don't have to worry about things like time and bankroll? The same applies to D'Alembert.

I think I have been able to calculate the EV for an infinite D'Alembert under the given stop conditions.
It appears that there are (2n)! / (n! (n+1)!) ways to have n wins and n losses such that the first result is a loss and losses > wins until the (2n)th bet.
If p is the probability of winning a particular bet, then the stop conditions are:
(a) the first bet is a win (probability p, result +1)
(b) the first two results are LW (probability (1-p) p, result +2)
(c) the first two results are LL, then n Ls and Ws that do not result in the number of Ws = the number of Ls (including the first 2) at any point, then WW (probability (2n)! / (n! (n+1)!) * pⁿ⁺² * (1-p)ⁿ⁺², result (n+2))

For a 50/50 game, this appears to diverge
For double-zero roulette, it appears to converge at around 5.210526
For a 25% game, it appears to converge at 0.625

Remember the key word - infinite. A finite D'Alembert has expected value zero for 50/50 and negative for a less than 50/50 game.

Quote: JyBrd0403

Just to finish the thought here. The LLN states that the average of the results (or win percentage) will become closer to the expected value the more trials performed. So, I did a little math experiment, starting at 10,000 trials with a win percentage of 49.1%, or -90 units from 0. I added a .001 percent increase to the win percentage every 10,000 trials. So, a modest move closer to 50% every 10,000 trials. Here's what I got.

10,000 trial intervals.

10k x .491 = 4,910 wins -90 from 0
20k x .492 = 9,840 wins -160 from 0
30k x .493 = 14,790 wins -210 from 0
40k x .494 = 19,760 wins -240 from 0
50k x .495 = 24,750 wins -250 from 0
60k x .496 = 29,760 wins -240 from 0
70k x .497 = 34,790 wins -210 from 0
80k x .498 = 39,840 wins -160 from 0
90k x .499 = 44,910 wins -90 from 0

As you can see, the longer you play the closer the win percentage gets to 50%, and the closer to 50% you get, the closer you get to 0.

Quote: ThatDonGuy

UPDATE #2: After some serious number crunching, I have managed to calculate the EV of a "true D'Alembert" (where you keep betting until you win with a minimum bet) for win probabilities from 0.4 to 0.499

Quote: JyBrd0403

Probability of winning is 50/50. Technically,the win rate wouldn't be used according to the LLN, but instead the average results every 10,000 trials. So, if the average results obtained means that you have to take the average wins every 10,000 trials. Then the average results would increase every 10,000 trials. Which....... kinda looks exactly the same.

1) 4,910 wins
2) 4,930 wins (avg. of 1,2) 4,920 wins .492
3) 4,950 wins (avg. of 1,2,3) 4,930 wins .493
4) 4,970 wins (avg. of 1,2,3,4) 4,940 wins .494
5) 4,990 wins (avg. of 1,2,3,4,5) 4,950 wins .495
6) 5,010 wins (avg. of 1,2,3,4,5,6) 4,960 wins .496
7) 5,030 wins (avg. of 1,2,3,4,5,6,7) 4,970 wins .497
8) 5,050 wins (avg. of 1,2,3,4,5,6,7,8) 4,980 wins .498
9) 5,070 wins (avg. of 1,2,3,4,5,6,7,8,9) 4,990 wins .499

The wave disappears doing it this way, but you can still do it the other way and see the wave :) But, it's pretty much the same thing. Which means that the wins move closer to the EV the more trials you run. The simulator results are still showing the exact opposite, the more trials the further from 0 you go. I'm sure if your simulator shows the avg. results it would have to be breaking the law of large numbers. I can't see any way the wins could keep straying away from 0 without breaking the law.

Quote: JyBrd0403

I don't understand your Table, are you saying for infinite bankroll and infinite time the EV is positive for the D'Alembert? Even at a .40 win probability?

Quote: ThatDonGuy

The fraction of wins approaches 0.5, but that does not necessarily mean that the wins minus losses value approaches zero - I have already shown a counterexample to that statement.
Here is the definition of the Law of Large Numbers from Wolfram Mathworld (emphasis mine):

A "law of large numbers" is one of several theorems expressing the idea that as the number of trials of a random process increases, the percentage difference between the expected and actual values goes to zero.


Yes - similar to the principle of "infinite monkeys and infinite typewriters," you "should" "eventually" catch up for any positive probability bet. It might take quadrillions of years, but it "should" happen "eventually." Of course, keep in mind that, under the same conditions, Martingale has positive EV as well.

I ran a 50/50 bet simulation using D'Alembert with the stop condition that you had to stop if you wiped out a bankroll of one billion (with a B) initial bets. At first, every sequence won, and the "per run" profit got as high as 6000 - then, about one billion runs into the simulation, it hit a bet that kept trying to crawl its way back to success but eventually hit the bankroll limit, and the per run profit (including all of the ones that led to the 6000 per run rate up to that point) was now a 6000 per run average loss. Note that the overall bet success was 49.99893%. This is in line with a general principle of pretty much every system - the fraction of wins may be high, but the ratio of average loss amount to average win amount makes up for it.

Quote: JyBrd0403

When you ran the simulation, did you see if the simulator broke the LLN? Did it go backwards, like. 49,9% to 49.8% to 49.7% to 49.3% etc. before finishing up at 49.99893%?

Quote: ThatDonGuy

However, LLN is not "set in stone"; if I play a 50/50 game a billion times and win 49.99999% of them, that does not mean that I have to win at least 49.99999% of the next billion as "otherwise that breaks LLN as my fraction of wins will go down." Believing otherwise is pretty much the definition of the Gambler's Fallacy.

Quote: JyBrd0403

Really? To defend the simulation results you're now going to throw out the Law of Large Numbers? Really?

Quote: DiscreteMaths2

I mean the betting system is literally named after Jean le Rond d'Alembert, the guy who thought the chance of a coin flip changed based on past results: https://www.cs.xu.edu/math/Sources/Dalembert/croix_ou_pile.pdf

There is no shame falling for Gambler's Fallacy, even an incredibly smart guy like d'Alembert did.

Quote: ThatDonGuy

What do you mean by "defend the simulation"? Are you claiming that, since the percentage of wins dropped at all once it got below 50%, then it is flawed because LLN says "that can't happen"?

I will repeat my earlier question:
How far down does the winning percentage have to go before you consider it to "break the LLN"?

Here's another question for you:
Suppose, in the first billion trials, there were 500,300,000 heads - a 50.03% win rate. Then, in the next billion trials, there were only 500,100,000 heads - a 50.01% win rate. The rate dropped from 50.03% to 50.02%, "as expected by LLN."
Now, switch the two sets of trials; the rate now increases from 50.01% in the first billion to 50.02% in the first two billion. This appears to contradict LLN.
Are you saying that the two sets of one billion events weren't equally as likely to be the "first set"?

Quote: DiscreteMaths2

All D'alembert does is modify the frequency distribution of the net gains / net losses at the end of a gaming session, which when you take all the sessions into account approaches the EV of the game after more and more sessions. Any other betting system you can think of approaches that same EV. If me and you bet on fair coin flips, the rate and manner we choose bet amounts isn't going to make the fair coin not 50/50. I mean the betting system is literally named after Jean le Rond d'Alembert, the guy who thought the chance of a coin flip changed based on past results: /math/Sources/Dalembert/croix_ou_pile.pdf

There is no shame falling for Gambler's Fallacy, even an incredibly smart guy like d'Alembert did.

Quote: unJon

Fun thread. Thanks for posting. What does your diverging formula say when p > 0.5 but the wins are less than double (so there’s still a HE). Such as banker bet in baccarat. Does formula not work?